Chapter 4, Problem 74QAP

A solution contains both iron(II) and iron(III) ions. A sample Of the solution is titrated with 35.0 ml, of M ${\text{KMnO}}_{\text{4}}$, which oxidizes ${\text{Fe}}^{2+}$ to ${\text{Fe}}^{3+}$. The permanganate ion is reduced to manganese(ll) ion. The equation for this reaction is

${\text{MnO}}_4{}^{-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)+5{\text{Fe}}^{2+}\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+5{\text{Fe}}^{3+}+4{\text{H}}_2\text{O}$Another 50.00-mL sample of the solution is treated with zinc, which reduces all the ${\text{Fe}}^{3+}$ to ${\text{Fe}}^{2+}$. The equation for this reaction is

$2{\text{Fe}}^{3+}\left(aq\right)+\text{Zn}\left(s\right)\to 2{\text{Fe}}^{2+}\left(aq\right)+{\text{Zn}}^{2+}\left(aq\right)$The resulting solution is again titrated with 0.0280 M ${\text{KMnO}}_{\text{4}}$; this time 48.0 ml, is required. What are the concentrations of ${\text{Fe}}^{2+}$ and ${\text{Fe}}^{3+}$ in the solution?

Interpretation Introduction

Interpretation:

The concentration of ${\text{Fe}}^{\text{2+}}$ and ${\text{Fe}}^{\text{3+}}$ in the solution should be calculated.

Concept introduction:

Number of moles is equal to the ratio of given mass to the molar mass.

The mathematical expression is given by:

Number of moles = $\frac{\text{given mass}}{\text{molar mass}}$

Molarity is defined as the ratio of number of moles of solute to the volume of the solution in liters.

The mathematical expression is given by:

Molarity = $\frac{\text{moles of solute }}{\text{volume of solution in liters}}$

A solution of salt of metal when reacts with other solution to give products, the formula which is used to find the volume of either solution is given by:

${\text{M}}_{\text{1}}{\text{V}}_{\text{1}}{\text{=M}}_{\text{2}}{\text{V}}_{\text{2}}$

Where, M₁ and M₂ are molarity of the solution

V₁ and V₂ are volume of the solution

Answer to Problem 74QAP

Concentration of ${\text{Fe}}^{\text{2+}}$ in sample is $98\times {10}^{-3}\text{ m}\mathrm{ole}/L$.

Concentration of ${\text{Fe}}^{\text{3+}}$ in sample is $36.4\times {10}^{-3}\text{ m}\mathrm{ole}/L$.

Explanation of Solution

Given information:

Volume of sample = 50.0 mL

Volume of ${\text{KMnO}}_4$ = 35.0 mL

Molarity of ${\text{KMnO}}_4$ = 0.0280 M

${\text{2MnO}}_{\text{4}}^{\text{-}}{\text{(aq)+8H}}^{\text{+}}{\text{(aq)+5Fe}}^{2+}\text{(aq)}\to {\text{Mn}}^{\text{2+}}{\text{(aq)+5Fe}}^{3+}{\text{(aq)+4H}}_{\text{2}}\text{O}$

${\text{2Fe}}^{3+}\text{(aq)+Zn(s)}\to {\text{2Fe}}^{2+}{\text{(aq)+Zn}}^{2+}(\text{aq)}$

The resulting solution titrated with:

Volume of ${\text{KMnO}}_4$ = 48.0 mL

Molarity of ${\text{KMnO}}_4$ = 0.0280 M

The given reaction is:

${\text{2MnO}}_{\text{4}}^{\text{-}}{\text{(aq)+8H}}^{\text{+}}{\text{(aq)+5Fe}}^{2+}\text{(aq)}\to {\text{Mn}}^{\text{2+}}{\text{(aq)+5Fe}}^{3+}{\text{(aq)+4H}}_{\text{2}}\text{O}$

Divide the above redox reaction into two half reactions that is oxidation half reaction and reduction half reaction.

Oxidation Half reaction:

${\text{Fe}}^{\text{2+}}\text{(aq)}\to {\text{Fe}}^{\text{3+}}\text{(aq)}$

Reduction half reaction:

${\text{MnO}}_4^{-}\text{(aq)}\to {\text{Mn}}^{\text{2+}}\text{(aq)}$

Assign oxidation states to each element in the above reactions.

Oxidation Half reaction:

${\text{Fe}}^{\text{2+}}\text{(aq)}\to {\text{Fe}}^{\text{3+}}\text{(aq)}$

Reduction half reaction:

$\stackrel{+7}{\text{Mn}}\stackrel{-2}{{\text{O}}_4^{-}}\text{(aq)}\to \stackrel{+2}{{\text{Mn}}^{\text{2+}}}\text{(aq)}$

Balance the elements except oxygen.

Oxidation Half reaction:

${\text{Fe}}^{\text{2+}}\text{(aq)}\to {\text{Fe}}^{\text{3+}}\text{(aq)}$

Reduction half reaction:

$\stackrel{+7}{\text{Mn}}\stackrel{-2}{{\text{O}}_4^{-}}\text{(aq)}\to \stackrel{+2}{{\text{Mn}}^{\text{2+}}}\text{(aq)}$

Now, add number of electrons (required) to that side of reaction where reduction occurs.

Oxidation Half reaction:

${\text{Fe}}^{\text{2+}}\text{(aq)}\to {\text{Fe}}^{\text{3+}}{\text{(aq)+e}}^{-}$

Reduction half reaction:

$\stackrel{+7}{\text{Mn}}\stackrel{-2}{{\text{O}}_4^{-}}{\text{(aq)+5e}}^{-}\to \stackrel{+2}{{\text{Mn}}^{\text{2+}}}\text{(aq)}$

Balance the charge by adding ${\text{H}}^{\text{+}}$ in both reactions.

Oxidation Half reaction:

${\text{Fe}}^{\text{2+}}\text{(aq)}\to {\text{Fe}}^{\text{3+}}{\text{(aq)+e}}^{-}$

Reduction half reaction:

$\stackrel{+7}{\text{Mn}}\stackrel{-2}{{\text{O}}_4^{-}}{\text{(aq)+5e}}^{-}+8{\text{H}}^{+}\to \stackrel{+2}{{\text{Mn}}^{\text{2+}}}\text{(aq)}$

Balance the hydrogen and oxygen atoms by adding water molecules.

Oxidation Half reaction:

${\text{Fe}}^{\text{2+}}\text{(aq)}\to {\text{Fe}}^{\text{3+}}{\text{(aq)+e}}^{-}$

Reduction half reaction:

$\stackrel{+7}{\text{Mn}}\stackrel{-2}{{\text{O}}_4^{-}}{\text{(aq)+5e}}^{-}+8{\text{H}}^{+}\to \stackrel{+2}{{\text{Mn}}^{\text{2+}}}{\text{(aq)+4H}}_{\text{2}}\text{O}$

Multiply the oxidation half reaction with 5 to balance the number of electrons in both reactions.

Oxidation Half reaction:

${\text{5Fe}}^{\text{2+}}\text{(aq)}\to 5{\text{Fe}}^{\text{3+}}{\text{(aq)+5e}}^{-}$

Reduction half reaction:

$\stackrel{+7}{\text{Mn}}\stackrel{-2}{{\text{O}}_4^{-}}{\text{(aq)+5e}}^{-}+8{\text{H}}^{+}\to \stackrel{+2}{{\text{Mn}}^{\text{2+}}}{\text{(aq)+4H}}_{\text{2}}\text{O}$

Now, add both reactions.

$\stackrel{+7}{\text{Mn}}\stackrel{-2}{{\text{O}}_4^{-}}{\text{(aq)+5e}}^{-}+8{\text{H}}^{+}\to \stackrel{+2}{{\text{Mn}}^{\text{2+}}}{\text{(aq)+4H}}_{\text{2}}\text{O}$

${\text{5Fe}}^{\text{2+}}\text{(aq)}\to 5{\text{Fe}}^{\text{3+}}{\text{(aq)+5e}}^{-}$

$\overline{\stackrel{}{{\text{MnO}}_4^{-}}\text{(aq)}+5{\text{Fe}}^{\text{2+}}({\text{aq)+8H}}^{\text{+}}\to \stackrel{}{{\text{Mn}}^{2+}}\text{(aq)+}5{\text{Fe}}^{\text{3+}}\text{(aq})+4{\text{H}}_2\text{O}}$

Now,

Molarity = $\frac{\text{moles of solute }}{\text{volume of solution in liters}}$

Convert the volume in mL to L.

Since, 1 L = 1000 mL

Thus, volume in L = $35.0\text{ mL}\times \frac{1\text{ L}}{1000\text{ mL}}=0.035\text{ L}$

Number of moles of ${\text{KMnO}}_4$ = Molarity × Volume of solution in liters

Number of moles = 0.0280 M× 0.035 L = $9.8\times {10}^{-4}\text{ mole}$

According to the reaction, the ratio between ${\text{KMnO}}_4$ and ${\text{Fe}}^{\text{2+}}$ is 1:5.

Thus, number of moles of ${\text{Fe}}^{\text{2+}}$ = $9.8\times {10}^{-4}\text{ mole}\times \text{5 = 4}\text{.9}\times {10}^{-3}\text{ mole}$

When the resulting solution titrated again:

Convert the volume in mL to L.

Since, 1 L = 1000 mL

Thus, volume in L = $\text{48 mL}\times \frac{1\text{ L}}{1000\text{ mL}}=0.048\text{ L}$

Number of moles of ${\text{KMnO}}_4$ = 0.0280 M× 0.048 L = $1.344\times {10}^{-3}\text{ mole}$

According to the reaction, the ratio between ${\text{KMnO}}_4$ and ${\text{Fe}}^{\text{2+}}$ is 1:5.

Thus, number of moles of ${\text{Fe}}^{\text{2+}}$ = $1.344\times {10}^{-3}\text{ mole}\times \text{5 = 6}\text{.72}\times {10}^{-3}\text{ mole}$

$\text{6}\text{.72}\times {10}^{-3}\text{ mole}$ is also equal to number of moles of ${\text{Fe}}^{\text{3+}}$ in 50.0 mL sample.

Number of moles of ${\text{Fe}}^{\text{3+}}$ = $\text{6}\text{.72}\times {10}^{-3}{\text{ mole of Fe}}^{2+}{\text{and Fe}}^{\text{3+}}-\text{4}\text{.9}\times {10}^{-3}{\text{ mole of Fe}}^{2+}\text{=1}\text{.82}\times {10}^{-3}\text{ mole}$

Concentration of ${\text{Fe}}^{\text{2+}}$ in sample = $\frac{\text{moles of solute }}{\text{volume of solution in liters}}$

Convert the volume in mL to L.

Since, 1 L = 1000 mL

Thus, volume in L = $50.0\text{ mL}\times \frac{1\text{ L}}{1000\text{ mL}}=0.05\text{ L}$

Concentration of ${\text{Fe}}^{\text{2+}}$ in sample = $\frac{\text{4}\text{.9}\times {10}^{-3}\text{ mole }}{\text{0}\text{.05 L}}=98\times {10}^{-3}\text{ m}\mathrm{ole}/L$

Concentration of ${\text{Fe}}^{\text{3+}}$ in sample = $\frac{\text{1}\text{.82}\times {10}^{-3}\text{ mole }}{\text{0}\text{.05 L}}=36.4\times {10}^{-3}\text{ m}\mathrm{ole}/L$

Conclusion

Concentration of ${\text{Fe}}^{\text{2+}}$ in sample is $98\times {10}^{-3}\text{ m}\mathrm{ole}/L$.

Concentration of ${\text{Fe}}^{\text{3+}}$ in sample is $36.4\times {10}^{-3}\text{ m}\mathrm{ole}/L$.